# Rate change

• July 3rd 2010, 07:24 PM
Wilmer
Rate change
Code:

```0 rate=9%=u%            1000.00=a   k rate=12%=v%   11=n                    3000.00=b```
Ok, here's the dope:
1000 bucks deposited in savings account at year 0, at 9% compounded annually.
12% compounded annually will later be the rate, effective at a time k which will
result in a future value of 3000 bucks after 11 years.
For this to happen, k = 5.451118.....

If opening rate = u and opening deposit = a,
second rate = v , future value = b and number of years = n,
(v > u, b > a(1+u)^n)
what is k in terms of a,b,n,u,v?
• July 4th 2010, 02:26 AM
CaptainBlack
Quote:

Originally Posted by Wilmer
Code:

```0 rate=9%=u%            1000.00=a   k rate=12%=v%   11=n                    3000.00=b```
Ok, here's the dope:
1000 bucks deposited in savings account at year 0, at 9% compounded annually.
12% compounded annually will later be the rate, effective at a time k which will
result in a future value of 3000 bucks after 11 years.
For this to happen, k = 5.451118.....

If opening rate = u and opening deposit = a,
second rate = v , future value = b and number of years = n,
(v > u, b > a(1+u)^n)
what is k in terms of a,b,n,u,v?

After $k$ years at rate $r_1$, and $(N-k)$ years at rate $r_2$ you have:

$b=a (1+r_1)^k(1+r_2)^{N-k}$

To solve for $k$ given $r_1$, $r_2$, $N$ and $a$ is an excercise in the laws of logarithms.

CB
• July 4th 2010, 04:14 AM
Wilmer
Yep...general case formula:

k = LOG[(1 + u) / (1 + v)] / LOG[b / a / (1 + v)^n]